Chapter 7.2, Problem 51E

Solution

a.

To determine: The coordinates of the rotated point in the $x\text{'}y\text{'}$ -coordinates system.

The obtained matrix is $\left[\begin{array}{cc}\frac{1+\sqrt{3}}{2}& \frac{-1+\sqrt{3}}{2}\end{array}\right]$ .

Given Information:

The given point of the coordinate in $xy$ plane is $\left(1,1\right)$ and the angle is rotated to 30 degree.

Explanation:

Consider the given matrix,

The points are defined as,

  $\begin{array}{c}\left[\begin{array}{cc}x\text{'}& y\text{'}\end{array}\right]=\left[\begin{array}{cc}x& y\end{array}\right]A\\ =\left[\begin{array}{cc}1& 1\end{array}\right]\left[\begin{array}{cc}\cos 30°& -\sin 30°\\ \sin 30°& \cos 30°\end{array}\right]\\ =\left[\begin{array}{cc}\left(\cos 30°\right)+1\left(\sin 30°\right)& 1\left(-\sin 30°\right)+1\cos 30°\end{array}\right]\\ =\left[\begin{array}{cc}\frac{1+\sqrt{3}}{2}& \frac{-1+\sqrt{3}}{2}\end{array}\right]\end{array}$

Therefore, the required matrix is $\left[\begin{array}{cc}\frac{1+\sqrt{3}}{2}& \frac{-1+\sqrt{3}}{2}\end{array}\right]$ .

b.

To determine: The coordinate of the point in the xy -coordinate system that was rotated to it.

The obtained matrix is $\left[\begin{array}{cc}\frac{\sqrt{3}-3}{2}& \frac{1+\sqrt{3}}{2}\end{array}\right]$ .

Given Information:

The given point of the coordinate in $xy$ plane is $\left(1,1\right)$ and the angle is rotated to 30 degree.

Explanation:

Consider the given information,

  $\begin{array}{l}x\text{'}=x\cos \alpha +y\sin \alpha \\ y\text{'}=-x\sin \alpha +y\cos \alpha \end{array}$

Substitute the given values in the above equations.

  $\begin{array}{l}1=x\cos 30°+y\sin 30°\\ 1=-x\sin 30°+y\cos 30°\end{array}$

And,

  $\begin{array}{c}1=\frac{\sqrt{3}x}{2}+\frac{y}{2}\\ \frac{1}{\sqrt{3}}=\frac{x}{2}+\frac{\sqrt{3}y}{2}\end{array}$

And,

  $1=-\frac{x}{2}+\frac{\sqrt{3}y}{2}$

Add the equation to solve the equation.

  $\begin{array}{c}1+\frac{1}{\sqrt{3}}=\frac{x}{2}+\frac{y}{2\sqrt{3}}-\frac{x}{2}+\frac{\sqrt{3}y}{2}\\ 1+\frac{1}{\sqrt{3}}=\frac{y}{2\sqrt{3}}+\frac{\sqrt{3}y}{2}\\ 1+\frac{1}{\sqrt{3}}=\frac{2y}{\sqrt{3}}\\ y=\frac{1+\sqrt{3}}{2}\end{array}$

Find the value of x .

  $\begin{array}{c}1=-\frac{x}{2}+\frac{\left(\frac{1+\sqrt{3}}{2}\right)}{2}\\ \frac{x}{2}=\frac{1+\sqrt{3}}{4}-1\\ x=\frac{1+\sqrt{3}-4}{2}\\ =\frac{\sqrt{3}-3}{2}\end{array}$

Therefore, the obtained matrix is $\left[\begin{array}{cc}\frac{\sqrt{3}-3}{2}& \frac{1+\sqrt{3}}{2}\end{array}\right]$ .